For the equation  $\frac{40}{x-1}-\frac{160}{x-4}-\frac{200}{x-5}+\frac{320}{x-8}=6x^2-27x,$  then which of the following option(s) is/are true?

Number of real solutions of above equation is 3

If E denotes the product of non-zero or complex roots of equation, then sum of divisors of  E is 2904

It  S  denotes the set of all real roots of equation then, sum of elements of  S  taken two at a time is 81

If  ${\alpha }_1,{\alpha }_2\in R$  be two roots of equation such that ${\log }_{{\alpha }_2}\left(2{\alpha }_1\right)$  is defined, then it must be 1